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150 





Class X^_^^A 

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Copyright N^ 



COPYRIGHT DEPOSnV 



MENSURATION 



FOR 



SHEET METAL WORKERS 



AS APPLIED IN WORKING ORDINARY 
PROBLEMS IN SHOP PRACTICE 



By WILLIAM NEUBECKER 



Reprinted from The Metal Wokkfr, Plumber anu Steam Fitter 



NEW YORK 

DAVID WILLIAMS COMPANY 

14-16 Park Place 



^^^1 



LIBRARY of OONeR^s| 
Two Copies ReceivdCi f 

JAN 28 ia08 

(}«t>yag<>t tnto 

0U8SA XXc, Nu. 

OOPY B. 



Copyright, 1907 
David Williams Company 



\ 



f-y^-f^. 






CONTENTS 



Page. 

Length of material required for a round pipe 6 

Amount of material for a square pipe 6 

Length of perimeter of a hexagon 6 

Square and circular sections of pipe of given dimen- 
sions 7 

Length of side of a square to be inclosed by a given 

circle 8 

Length of arc when angle and radius are known g 

Amount of material for an ellipse when length and width 

are given 9 

Length of hypotenuse in a right angle triangle 9 

Areas in square, rectangle and rhomboid lO 

Areas in a right and an oblique triangle 1 1 

Area of triangle when three sides are given I2 

Area of trapezoid 13 

Area of trapezium 14 

Area of circle 14 

Area in circle and square 14 

Area in a ring 15 



2 Mensuration for Sheet Metal Workers. 

Page. 
Areas in sector and segment of a circle 15 

Square whose area is equal to area of given circle 16 

Circle whose area is equal to area of given square 17 

Area of ellipse 17 

Area of regular polygon 18 

Area of sphere 18 

Convex surface of cylinder 18 

Convex surface of frustum of cylinder 19 

Convex surface of elliptical cylinder and frustum of 

elliptical cylinder 19 

Convex surface of right prism and of frustum of right 

prism 20 

Convex surface of right cone 20 

Convex surface of right pyramid 21 

Convex surface of frustum of right cone 21 

Convex surface of frustum of right pyramid 22 

Capacities of tanks, etc 22 

Contents of cube 23 

Contents of hexagonal prism 24 

Contents of c)'linder 24 

Contents of sphere 25 

Contents of cone 25 

Contents of pyramid 25 

Contents of hopper 2^ 

Contents of prism of pyramid 27 



Contents. 3 

Page. 
Contents of frustum of cone 29 

Contents of prismoid , 29 

Contents of wedge 30 

Practical examples for the shop 30 

Pipes equal to rectangular pipe 30 

Boot 31 

Chimney top , 31 

Another form of boot 32 

Pipe with branches 33 

Triangular pipe 34 

Elliptical pipe 34 

Two pronged fork 35 

Three pronged fork 35 

Ascertaining the sizes of articles 36 

A square tank 36 

A sphere 37 

An oil tank 39 

Wash boiler 39 

Flaring pail 40 

Flaring measure 40 

Flaring elliptical tub 41 

Flaring pan 41 

Short rules in computation 42 

Diameter of circle 42 

Height of inaccessible point 42 



4 Mensuration for Sheet Metal Workers. 

Page. 
Use of steel square 44 

Problems of steel square solution 44 

Obtaining radius 45 

Rectangle to square 47 

Circle to square 47 

Three squares . .47 

Circle of four times the area 48 

Circumference of an ellipse 48 



1 



Mensuration for Sheet Metal Workers, 



Very little has been written about mensuration for 
the sheet metal worker, and although various collections 
of tables have been published, the rules are not generally 
explained so as to be easily understood or to enable one 
to proceed intelligently with practical problems which 
come up. In the pages which follow examples in com- 
puting the circumferences, areas and capacities for va- 
rious shapes arising in practice are given in detail. The 
comprehension of these will enable the student to com- 
pute any ordinary problem in the shop. 

Besides methods for finding the lengths, areas and 
volumes of the simpler geometrical forms, examples are 
given in computing the areas of heating and ventilating 
pipes of all ordinary shapes, making their areas equal to 
those of pipes of other profiles. The use of the pris- 
moidal formula for obtaining the capacities of various 
shaped bodies is fully explained, as is a method of ob- 
taining the hight of any solid to hold a given quantity 
when the diameter is known, or vice versa. A short rule 
is illustrated for finding the diameters of branch pipes 
taken from a given main pipe, so that the areas of the 
branches will equal the area of the main, and many other 
problems are treated. 

One of the first problems arising in the shop is to find 
the true length of material required for round or other 
shaped pipes. The rule for obtaining the circumference 



6 Mensuration for Sheet Metal Workers. 

of any circle is to multiply the diameter by 3.1416, or, 
as is sometimes used in practice, by 3 1-7. 

In Fig I, A represents a circle 4 inches in diameter. 
Then 4 inches X 3-i4i6 ^ 12.5664, or 12 9-16 inches 




CIRCUMFERENCE 



12.5664 5 

Fig. 1. — Finding Length of Matorial Required for a Round Pipe. 

approximately, the circumference, as shown rolled out 
from a to b. 

Referring to Fig. 2, let abed represent a 4-inch 
square. To obtain the perimeter or amount of material 
for this pipe, it is only necessary to multiply 4 inches by 
4. which equals 16 inches. As shown from a to a', the 
distance represents the length of the sum of the sides of 
the square figure. 

In the same manner the length of the perimeter of 
the hexagon is obtained in Fig. 3. In this each side meas- 



o 



A^ PERIMETER 



J I 



d *" a 5' c' d' «' 

Fig. 2. — Finding Amount of Material for a Squai-e Pipe. 

ures- 3 inches. The figure has six sides, so we have 6 X 
3 inches = 18 inches, the length shown from a to a'. 

Sometimes a round pi])e is to be formed to a square 
section at the opposite end, using the same amount of 
material as in the round pipe. This makes it desirable 



Mensuration for Sheet Metal Workers. 7 

to know what the length of the sides at the square end 
will be. Knowing that the diameter of the circle a in 
Fig. 4 is 4 inches, and that the circumference is 12.5664, 

d 3" c 



PERIMETER 



6' c d' e' f 

Fig. 3. — Finding Length of Perimeter of a Hexagon. 



it is only necessary to divide by four, which will give 
3.1416 inches, the length of the sides of the square b. If 
the circumference is not known, multiply the diameter by 
0.7854; thus, 4 inches X 0.7854 = 3.1416 inches, which 




3.h41C 




^'^vJBF^^' 




Figs. 4 and 5.— Finding Square and Cir- 
cular Sections of Pipe of Given Dimen- 
sions. 



T'ig. 6.— Finding Length 
of Side of a Square to 
Be Inclosed by a Given 
Circle. 



multiplied by four sides = 12.5664 inches, the perimeter 
for the square b, and is equal to the circumference of the 
circle a, proving the above rules. 



8 Mensuration for Sheet Metal Workers. 

If, however, the conditions were reversed, and each 
side of a given square measured 5 inches, as shown at a 
in Fig. 5, making the perimeter of the square 20 inches, 
and it is desired to know what diameter a circle would 
have whose circumference would be equal to that peri- 
meter, to obtain this diameter multiply the length of one 
side, or 5 inches, by 1.2732, which equals 6.366 inches, 
the required diameter. ]\Iultiply this diameter,. 6.366 




Fig. 7. — Finding Length of 
Arc Wlien Angle and 
Radius Are Known. 



Fig. 8. — Finding Amount 
of Material for an I'll ipse 
When Length and Width 
Are Given. 



Fi 



9.— Find- 
ing Length of 
Hypotenuse in 
a Right Angle 
Triangle. 



inches by 3.1416, and the product will be 20 inches, the 
perimeter of the square thus proving the rule. 

When double ventilation pipes are constructed, as 
shown in Fig. 6, where the outer pipe is a true circle and 
the inner pipe a square, allowing an air space between 
the two pipes, and it is desired to know the length of the 
side of a square to pass inside of a given circle, it is only 
necessary to multiply the diameter of the given pipe by 
0.7071. Suppose the round pipe a were 6 inches in diam- 
eter, then 6 inches X 0.7071 = 4,2426, or nearly 4j4 
inches, for the side of the desired square. 

When it is desired to find the lengfth of an arc, when 



Mensuration for Sheet Metal JVorkers. 



9 



only the angle and radius are known, then multiply the 
number of degrees by the diameter and the product by 
0.008727. In Fig. 7 it is desired to find the length of the 
90-degree arc a b, whose radius is 4.5 inches, or diameter 
9 inches. Following the above rule, we have 90 X 9 = 
810. 810 X 0.008727 = 7.0688 inches, the length of a b. 
When the length and width of an ellipse are given, 
and it is desired to know how much material is required 
for its circumference, multiply half the sum of the two 
diameters by 3.1416. Thus, in Fig. 8, we have an ellipse 



be m 



Area 

81 sq. inches 



lA 



Area 



/ 130,5 sq, inches 



H ^9= sj i*-- ]Ai—- 



~1 



Fig. 10. — Areas in Square, Rectangle and Rhomboid. 

the major axis of which is 10 inches and the minor axis 

10 + 6 



6 inches. Following the above rule, 



= 8 X 



3.1416 = 25.1328. It should be understood that the cir- 
cumference of an ellipse cannot be accurately determined, 
and the above rule is merely an approximation giving 
fairly close results. 

When a large smoke stack is to be carried up at an 
angle on a building and the vertical hight and horizontal 
projection are known, the length of the slant can be ob- 
tained by the rule illustrated in Fig. 9, which shows how 
the length of the hypotenuse is found in a right angle tri- 
angle. Add the square of the base to the square of the 
altitude, and the square root of the sum will be the hypo- 
tenuse. The base being 6 inches and the altitude 8 inches. 



10 Mensuration for Sheet Metal JVorkers. 



we have V6- -|- 8" = V36 -|- 64 = Vioo = 10. The 
square root of 100 is 10, because 10 is the number which, 
when multiphed by itself, will equal 100. 

The area of any surface is the number of square 
inches or square feet within its outline. In connection 
with the illustrations, which immediately follow, the rules 
will be given for obtaining the areas of the various geo- 
metrical shapes, which will enable the student to proceed 
intelligently when computing areas and capacities of va- 
rious forms arising in practice. While many sheet metal 
workers understand how to compute the areas of the 
more common geometrical shapes, there are some who do 




Fig. 11. — Areas in a Right and an Oljlique Triangle. 

not know how this branch of mathematics can be applied 
in practical work. 

One of the most simple figures, the area of which the 
student usually masters first, is that of the square, shown 
hy a b c d in Fig. 10, each side of which measures 9 
inches. To obtain the area simply multiply the length 
of the side by itself; thus: 9 X 9 = 81 square inches. 

To obtain the area of the rectangle e f h i, multiply 
the width by the length, thus: 9 X i4-5 = 130.5 square 
inches. 

Suppose a surface had to be covered with sheet metal, 
the shape being that of a rhomboid, shown by ni n h i, 



Mensuration for Sheet Metal Workers. 11 

in which m n is parallel to and of the same length as h i; 
then knowing the perpendicular hight e i to be 9 inches, 
and the length to be 14^^ inches, we would have the same 
area as that shown in the rectangle e f h i, because the 
triangle / 11 h equals the triangle e m i. 

When the area of a triangle is to be found, whether 
right or oblique, the base and perpendicular hight be- 
ing known, the rule is to multiply the perpendicular 
hight by the base, and half the product is the area. In 
Fig. II, let a b c represent a right triangle, whose base is 
8 inches and hight 12 inches; then 12 X 8 = 96. 96 -f- 




Flg. 12. — Area of Triangle When Three Si(le.s Are Given. 

2 = 48 square inches, the area. If the product 96 were 
not divided by two it would represent the area of the rec- 
tangle a i c b, but by drawing the diagonal a c we divide 
the rectangle into two right triangles, each equal to one- 
half of 96, or 48, as shown. 

The diagram e d f represents an oblique triangle, 
whose base is 15 inches and perpendicular hight to the 
apex 12 inches. Now, following the above rule, we have 
12 X 15 = 180. 180 -^ 2 = 90 square inches area of 
d e f. To prove this, construct the rectangle e in n f, 
and from d drop the vertical line d o. The distances ni d 
and d n are 9 and 6 inches, respectively. We then have 
two rectangles, one 9 x 12 and the other 6 x 12 inches. 



12 



Mciisiiratidii for SJicct Metal JVorkers. 



By drawing d e and d f we obtain two right triangles, 

doc and d o f. Following the explanation given in con- 

•^1 7 u 9 X 12 , 6 X 12 

nection with a b c, we have = S4) and 

2 ^^' 2 

= 36 ; 36 -L- 54 = 90 square inches, the same as the area 
oi d e f. 

Sometimes an irregular shaped structure is to be cov- 
ered and none of the angles and only the dimensions of 
the three sides are known or can be obtained. The rule 
to be followed in this case is as follows : From half the 




1^ 12'-0: ^ 

Fig. 115. — Area of Trapezoid. 




i'.O' 
Fig. 14. — Area of Trapezium. 



sum of the three sides subtract each side separately ; find 
the continued product of the half sum of the sides and 
the three remainders, and the square root of this product 
is equal to the area of the triangle. In Fig. 12 is shown 
a portion of a surface, each side measuring respectively 
21, 17 and 10 feet. The sum of the three sides is 21 -{- 
17 -f- 10 = 48. 48 ^ 2 = 24, the half sum. Subtract- 
ing each side separately from half the sum of the sides, 
we get the three remainders, 24 — 21 ^3 feet; 24 — 
17 = 7 feet, and 24 — 10 = 14 feet. Now, 24 X 3 X 7 
X 14 = 7056. V7056 = 84. 

The method of obtaining the square root is as fol- 



Mensuration for Sheet Metal Workers. 13 

lows : Pointing off the number 7056 into periods of two 
figures each, we get 7o'^6, which shows that the complete 
part of the square root contains two figures. Then pro- 
ceed as below : 

Root. 
Trial divisor. Correct divisor. 70'.o6(84 ans. 

160 164 64 

656 
656 

It will be observed that the greatest number whose 
square is contained in 70 is 8 ; and, therefore, 8 is the 
first root of the figure; 8 X 8 = 64. Subtracting 64 
from 70 and bringing down the next period, 56, we get 
the first partial dividend, 656. The double of 8, the 
partial root already found, is 16, and annexing a cipher to 
this we get 160 as the first trial divisor. This trial di- 
visor is contained in the partial dividend 656 four times, 
which suggests four as the second figure of the root. 
Adding 4 to 160 we obtain 164 as the correct divisor. 
When the product (4 X 164 = 656) is subtracted from 
the partial dividend 656 there is no remainder. Eighty- 
four is the required square root, and 84 X 84 =: 7056- 
The triangle in Fig. 12 contains 84 square feet area. 

In Fig. 13 is shown a trapezoid, and in Fig. 14 a 
trapezium. The difference between the two lies in the 
fact that the trapezoid has two sides parallel to each 
other, while the trapezium has no sides parallel. When 
computing the area of an irregular surface having two 
parallel sides use the rule for obtaining the area of a 
trapezoid, shown in Fig. 13, which is as follows: Mul- 
tiply half the sum of the parallel sides by the perpendic- 
ular hight. Half the sum of the parallel sides in Fig. 
1 2 is 6 + 12 = 18 ; 18 -f- 2 = 9, the mean distance. Then 



li 



Mensuration for Sheet Metal JVorkers. 



9X5 (the perpendicular higlit) = 45 square feet area in 
abed. 

In computing the area oi a d e c, in Fig. 14, we di- 
vide it into two triangles and a trapezoid, by drawing 
vertical lines from the angles a and d, at right angles to 
c c. The bases of the two triangles are 4 and 2 feet, re- 



FULL SIZE 




Fig. 15. — Area in Circle. 




Fig. 16. — Area in Circle and Square. 



spectively, and the altitudes 8 and 12 feet, respectively ; 

.1 ,• r , 8 + 12 _ 20 

the mean distance of the trapezoid is — =10 



Then 



8X4 ^ 32 ^ J 



6, area of triangle a b c; 



24 
— — =12, area of triangle d e f; 10 X 7 



12X2 
2 

70, area 

of trapezoid b a d f. Then 16 -f 12 -f 70 = 98 square 
feet area m a d e c. 

To obtain the area of a circle, square the diameter 
and multiply by 0.7854. Fig. 15 shows a section of a 
pipe ioy2 inches in diameter, of which it is desired to 
find the area. Then, 10.5 X 10.5 = 110.25; 110.25 X 
0-7854 = 86.59 square inches. Fig. 16 shows a square 
containing i square inch. The inscribed circle is i inch 
in diameter. Although it is i inch in diameter, it con- 
tains only 0.7854 square inch. 



Mensuration for Slicct Metal JJ^orkers. 



15 



When the area of a ring is to be determined, as shown 
in Fig. 17. in which the outside diameter is 12 inches and 
the inside diameter 5 inches, deduct the square of the 
small diameter from the square of the large diameter and 
m.ultiply the remainder by 0.7854. Following this rule, 
we have 12- — 5- = 12 X 12 — 5 X 5 = i44 — 25 = 
119; 119 X 0.7854 =: 93.462 square inches in the ring. 

Sometimes a ventilating pipe is to be constructed, 
whose section is a sector of a circle, as shown hy a b c e 
in Fig. 18, and it is necessary to know its area. In the 





AREA OF 

SEGMENT 

6. 136 

8Q. INCHES 



Fig. 17. — Area in a Ring. 



Pig. 18.- 



-Areas in Sector and Segment 
of Circle. 



illustration, the radius ^ c is 3^ inches; the angle a e c, 
105 degrees (which is determined by the use of the pro- 
tractor) ; the chord a c, 6 inches, and the rise or middle 
ordinate m b, i}^ inches. Before obtaining the area, it 
will be necessary to know the length of the arc a c, which 
is obtained by following the rule given in connection with 
Fig. 7, Part I, as follows: In Fig. 18 we have 105 X 7-5 
X 0.008727 = 6.872 inches, the desired length of arc. 
Then to obtain the area oi a b c c, multiply the length of 
the arc by half the radius. The radius is 3.75 ; 3.75 -^ 
2 = 1.875. Then 1.875 X 6.872 = 12.885, the area of 
the sector. 



16 



Mcnsurafioii for Sheet Metal Workers. 



If the area of the segment a b c a is desired, knowing 
the area of the sector, it is only necessary to deduct from 
this amount the area of the triangle a e e. Thus half the 
distance of the chord a c is 3 inches, and the hight from 
the chord in to the center e is 2}i inches. Then 3 X 
2.25 = 6.75 square inches, the area of a c ^. Then 12.885 
(the area of the sector) — 6.75 (the area of the triangle) 
= 6.135 square inches, the area of the segment. 

If the chord a c and the rise m b of the segment were 




K-- — 6 i 



80. INCHES 

c 



AREA 

78.64 

8Q. INCHES 

B 



K — 8t862- 





Fig. 19.— Square Whose Fig. 20. — Circle Whose 
Area Is Equal to Area Area Is Equal to Area 
of Given Circle. of Given Square. 



Fig. 21. — Area 
of Ellipse. 



given, and it was desired to find the diameter of the 
circle of which the segment is a part, divide the sum of 
the squares of half the chord and the rise by the rise, 
and the quotient is the desired diameter. Half of the 
chord a c is 3 inches, and the rise m b is lYi inches. Fol- 
lowing the above rule, we have 3^ -f- 1.5- = 11.25; H-^S 
-^ 1.5 = 7.5 inches, the diameter. As twice the radius 
e c equals 73^ inches, the above rule is proven. 

It sometimes happens that a transition is to be made 
for a heating or ventilating pipe from round to square, 



Mensuration for Sheet Metal JVorkers. 17 

and the square end is to have the same area as the given 
round end. Let A in Fig. 19 represent a section of a 10- 
inch heating pipe ; then, following the rule given in con- 
nection with Fig. 15, the area of A in Fig. 19 is 78.54 
square inches. To find the dimensions of a square of 
equal area, multiply the given diameter by 0.8862. Thus 
10 X 0.8862 = 8.862, or the side of the square B. 

If, however, the conditions were reversed and the 
square end of the transition piece were given, say 6 
inches, as shown in C, Fig. 20, then multiply the given 
side by 1.1284, and the product will be the diameter of a 





i-.-IO-- 

Pig. 22. — Area in Regular Polygon. Fig. 2.3. — Area of Sphere. 

circle of equal area. Thus 6 X 1.1284 = 6.7704 inches, 
diameter of circle D. 

When the area of an ellipse is required, multiply the 
long diameter by the short diameter, and this product by 
0.7854. In Fig. 21 is shown an ellipse, whose long diam- 
eter is 14 inches and short diameter 8 inches. Then 8 X 
14 = 112; 112 X 0.7854 = 87.96 square inches area. 
The method for obtaining the dimensions of the opposite 
end of a transition piece, with one end a given ellipse and 
the opposite end to have similar area in either round, 
square or rectangular section, will be explained later un- 
der the head of Practical Examples for the Shop. 

When a surface is to be covered with sheet metal 
whose shape is that of any regular polygon, the rule to 



18 



Mensuration for Sheet Metal JJ'orkcrs. 



be followed in obtaining the area is to multiply the sum 
of the sides by half the perpendicular distance from cen- 
ter to sides. For example, there is shown in Fig. 22 a 
regular polygon having six sides, called a hexagon. The 
one side, c d, equals 10 inches, and the sum of the six 
sides, 6 X 10, or 60 inches. The whole perpendicular 
distance & a is 8.66 inches, which divided by two gives 
4.33, or half the perpendicular distance. Then 60 X 4-33 
= 259.8 square inches area in the hexagon. 

When a sphere is to be made of copper or sheet iron, 



CONVEX 
SURFACE 



..I 





Fig.24. — Convex Surface of 

Cylinder. 



Fig. 25. — Convex Surface of 
Frustum of Cylinder. 



and it is required to know the amount of material it will 
take, the area is obtained by squaring the diameter of the 
sphere and multiplying by 3.1416. The sphere shown in 
Fig. 23 is 12 inches in diameter. To find its area, we 
have 12 X 12 = 144; 144 X 3.1416 = 452.39 square 
inches area. 

In Fig. 24, A is the plan of a cylinder 10 inches in 
diameter and B the elevation, 12 inches high. It is de- 
sired to find the convex surface of this cylinder. In this 
problem, as well as in others which will follow, the areas 
of the ends will not be considered, as this was explained 
in previous problems. The rule to be employed in obtain- 



Mensuration for Sheet Metal Workers. 



19 



ing the area of the convex surface of any cylinder is to 

multiply the circumference by the hight. As the cylinder 

is 10 inches in diameter and the hight 12 inches, then 10 

X 3-1416 := 31.416, or circumference; 31.416 X 12 = 

376.992 square inches in convex surface. 

In Fig. 25, A and B show the plan and elevation of a 

frustum of a cylinder. To obtain this convex surface, 

multiply one-half the sum of the greatest and least bights 

by the circumference. The greatest hight is 18, the least 

28 
10; 10 + 18 = 28; = 14; 14 X 3-1416 X 12 = 

527.78 square inches area. 



CONVEX 

SURFACE 

1006,31 SQ.IN, 





CONVEX > 


-r 


SURFACE ^ 


703. 


''.< B' 




SQ.IN 


^\ 


r 


T'T 




J. 


J- 


1 










Fig. 26. — Convex Surface of Elliptical Cylinder and Frustum of 
Elliptical Cylinder. 



Another problem often arising in the shop is to find 
the convex surface of an elliptical tank. The same rule 
is employed as in a cylinder — that is, multiply the cir- 
cumference by the hight. In Fig. 26, A and B and A^ 
and B^ show, respectively, the plans and elevations of an 
elliptical cylinder and the frustum of an elliptical cylin- 
der, whose long diameter is 20 inches and short diameter 
12 inches. Following the rule given in Fig. 8, Part I, 



for obtaining the circumference, we have 



20 -h 12 



32 



= 16; 16 X 3-1416 = 50-2656; 50-2656 X 20 = 1005.31 
square inches in the convex surface B. For the frustum 



20 



Mensuration for Sheet Metal Workers. 



B^ we have 



^^ = 14; 14 X 50.2656 (the cir 



+ 20 28 
2 
cumference) = 703.71 square inches. 

The same rules apphcable to the cyUnder are also ap- 
plicable to prisms whose bases are regular polygons, 
whether they be right prisms or frustums. In Fig. 2y, 
A and B represent, respectively, the plan and elevation 
of a prism 18 inches high, each side of the polygon being 



T' 




i 




•1 
T 




c.s. 

643 

ig.iN. 




_l. 


_ 




_ 





MH 



Fig. 27. — Convex Surface of Right 
Prism and of Frustum of Right 
Prism. 



1 c.s. \ 

• / 339.29 
// SQ.fN. 


\ 


1 

1 

Y 

L ,2— - 




\A^ 


J 


"'ig. 28. — Convex 


Surface of 


Right 


Cone. 





6 inches. The perimeter of the hexagon is 6 X 6 = 36 
inches ; 36 X 18 = 648 square inches in the convex sur- 
face B. For the convex surface of the frustum, we have 
10 -f 18 _ 28 
2 2 



14; 14 X Z^ (the perimeter) = 504 



square inches. 

When the convex surface of any right cone or pyra- 
mid is desired, then multiply the circumference, or pe- 
riphery, of base by half the slant bight. In Fig. 28 A 
and B show the plan and elevation of a right cone, which 
will serve as an example. The diameter of the cone at 
its base is 12 inches; its circumference is therefore 12 



Mensuration for Sheet Metal JJ^orkers. 



21 



X 3-i4i6 = 37-6992; half of the slant hight is 9; then 
9 X 37-6992 = 339-29 square inches in the convex sur- 
face of the cone B. 

Using the same rule for Fig. 29, whose base or plan 
A is a hexagon with 7-inch sides, we have 7 X 6 = 42 ; 
9 X 42 = 378 square inches in the convex surface of the 
pyramid B. 

Suppose the elevations in Figs. 28 and 29 were cut 
off parallel to the base, forming the frustums of a cone 




Fig. 29. — Convex 
Surface of Right 
Pyramid. 



Fig. 30. — Convex Sur- Fig. 31. — Convex Sur- 
face of Frustum of face of Frustum of 
Righit Cone. Right Pyramid. 



and pyramid. The rule for obtaining the area of these 
convex surfaces is to multiply the circumferences or pe- 
ripheries of the two ends by half the slant hight. An 
example of this problem is given in Fig. 30, in which the 
plan of the base A is 10 inches and the diameter at the 
top 5 inches. Then 5 X 3-i4i6 == 15.708, or circum- 
ference at top, and 10 X 3-1416 = 31 -416, or circum- 
ference at base; 15.708 + 31.416 = 47.124; 47.124 X 5 
(half the slant hight) = 235.62 square inches of convex 
surface. To prove this problem, we will assume that we 
have a right cone in Fig. 30, whose base is 10 inches and 



22 



Mensuration for Sheet Metal JVorkers. 



slant hight 20 inches. Then, following the rule given in 
connection with Fig. 28, we have 10 X 3-i4i6 X 10 = 
314.16 square inches convex surface in whole cone. The 
area of the upper half of the cone shown by dotted lines, 
whose base is 5 inches, is 5 X 3-i4i6 X 5 = 78-54 square 
inches; 314.16 — 78.54 = 235.62 square inches, the area 
of the frustum of the cone, proving the problem. 

In Fig. 31, A is a regular polygon, with 6-inch sides, 
and the plan of the base of the frustum of a pyramid; 




Fig. 32. — Contents of Cube. 



J 


i 


\ 


CONTENTS = 




15,588 CU.IN.. 









OR 
67,48 GALLONS 


i 


>-■■ 




jL- 


\ 


w 






—10"^ 




Fig. 33.— Contents of Hexagonal 






Prism 





B the plan of the top, whose sides are 3 inches ; C, the 
elevation, has a slant hight of 8 inches. Then 6X6 = 
36, or perimeter of base; 3 X 6 = 18, or perimeter of 
top; 36 -|- 18 = 54; one-half the slant hight is 4; 4 X 
54 =: 216 square inches in C. 



Finding Capacities of Tanks, Etc. 

Some sheet metal workers would be at a loss to pro- 
ceed if a customer came in the shop and required a tank 
constructed of No. 20 galvanized sheet iron to hold 63^ 
gallons, the tank to fit in a space 51 inches high. Know- 
ing the number of gallons and the hight, it would be 
necessary to know the diameter before the tank could be 



Mensuration for Sheet Metal Workers. 23 

laid out. While rules given in various publications are 
understood by those versed in mensuration, the less 
skilled do not know how to apply them practically. In 
computing the capacity of any vessel we deal with cubic 
and liquid measures, and therefore it may not be out of 
place to present the tables of cubic and liquid measures : 

Cubic or Solid Measure. 
1728 cubic inches = 1 cubic foot, or 12 x 12 x 12. 
27 cubic feet — 1 cubic yard, or 3 x 3 x 3. 
231 cubic inclies = 1 United States gallon. 
57.75 cubic inches = 1 United States quart. 
28.875 cubic inches = 1 United States pint. 
7.21875 cubic inches = 1 United States gill. 

Liquid Measure. 
4 gills = 1 pint. 
2 pints = 1 quart. 
4 quarts = 1 gallon. 
31% gallons = 1 barrel. 
63 gallons, or 2 barrels = 1 hogshead. 
1 gallon ■— 4 quarts — 8 pints = 32 gills. 

In the problems that follow practical exam.ples have 
been given as they are apt to arise in the shop, and the 
student should have no difficulty in learning to figure the 
capacity of any vessel which may arise in practice. One 
of the most simple forms to be computed is that of a cube 
or square tank, shown in Fig. 32. Here we have a tank 
8x8 feet square and 8 feet high. The rule to be em- 
ployed in finding the solidity, whether the base is a square 
or rectangle, is to multiply the length of any one side by 
its adjoining side and multiply the product obtained by 
the hight. Then we have 8 X 8 r= 64; 64 X 8 = 512 
cubic feet. As a cubic foot contains 1728 cubic inches, 
we have 1728 X 512, or 884,736 cubic inches. To find 
the number of gallons the tank will hold divide 884,736 
by 231, the number of cubic inches in a gallon, and we 
get 884,736 -^ 231 = 3830 gallons, and 6 cubic inches 
over. 



24 



Mensuration for Sheet Metal JVorkers. 



Suppose a tank is to be constructed whose base is any 
regular polygon, as shown in Fig. 33, where a tank is 
shown 5 feet high whose base is a hexagon, each side 
measuring 10 inches. The rule to be used is to multiply 
the area of the base by the hight. Referring to Fig. 22, 
;we find that the area of a hexagon whose side is 10 
inches is 259.8 square inches, which multiplied by the 
hight in Fig. 33 will give the cubic contents. The hight 
shown is 5 feet. As the area of the base is in inches, 
then reduce the hight to inches; then 60 X 259.8 = 15,- 




-1_ 

Fig. 34.— Contents of Cylinder. 




CONTENTS = 

U,137.A 

CU..(N. 

OR 

61.2 GALLONS 



Fig. 35. — Contents of Spliere. 



588 cubic inches. For the number of gallons, 15,588 ~ 
231 =: 67^ gallons, scant. 

In Fig. 34 is shown a cyclinder or round tank, the con- 
tents of which is obtained by the same rule as in the pre- 
ceding figure. The bottom of this round tank is 10 inches 
in diameter and its hight 303^ inches. Then 10- X 0.7854 
equals the area of the base; or 10 X 10= 100; 100 X 
0.7854 = 78.54 square inches ; 30.5 X 78-54 — 239547 
cubic inches. For the number of gallons, divide the above 
product by 231 and the quotient will be 10.37 gallons. 

When a copper ball is made to use as a float in a large 
tank, it is sometimes desirable to know the number of 



Mensuration for Sheet Metal Workers. 



25 



cubic inches in same. Thus in Fig. 35 is shown a sphere 
30 inches in diameter, whose capacity is formed by muhi- 
plying the cube of the diameter by 0.5236, or 30'"' = 2/,- 
000; 27,000X0.5236=14,137.2 cubic inches; 14. 137.2 
-^231 =61.2 gallons capacity. 

When the contents are required of a cone or pyramid, 
the rule to follow is to multiply the area of the base by 
one-third the perpendicular hight. In Fig. 36 is shown 1 



CONTENTS = 
5,026.66 CO. IN. 



21.76 GALLONS 




Fig. 36. — Contents of Cone. 




CONTENTS = 
2,598 CU.IN. 



Il-Z't GALLONS 



Fig. 37. — Contents of Pyramid. 



cone, whose base is 20 inches and whose vertical hight is 
48 inches. Following the above rule, we have 20 X 20 = 
400; 400 X 0-7854 = 314.16 square inches, or area, mul- 
tiplying which by one-third the vertical hight, 48, or 16, 
we have 314.16 X 16 = 5026.56 cubic inches. Dividing 
this by 231, the number of cubic inches in a gallon, we 
get 21.76 gallons. 

The method of computing the solidity of a pyramid is 
shown in Fig. ^7, which shows a pyramid whose base is a 
hexagon, each side being 10 inches, and whose vertical 
hight is 30 inches. As the area of a hexagon each side 
of which is 10 inches is 259.8 square inches, then multiply 



26 



Mensuration for Sheet Metal Workers. 



this amount by one-third the vertical hight (30), or 259.S 
X 10 ^ 2598 cubic inches, which divide by 231 and we 
get 11.24 gallons. 

The prismoidal formula can be used in calculating the 
volume or capacity of a prism, cylinder, cone, pyramid, 
frustum of a cone or pyramid, wedge, as well as many 
irregular shaped bodies. A prismoid is by definition a 
solid whose bases are polygons and lie in parallel plans 
and whose faces are quadrilaterals or tri-angles. To find 




Fig. 38. — Contents of Hopper. 




Fig. 30. — Contents of Frustum 
of Pyramid. 



the contents of a prismoid, or any of the above mentioned 
solids, add together the areas of the two parallel planes 
and four times the area of a section taken midw^ay be- 
tween and parallel to them, and multiply the sum by one- 
sixth of the perpendicular distance between the parallel 
planes. 

Applying this rule for obtaining the volume of a 
wedge, pyramid or cone, the area of the upper base is o, 
because it runs to an apex. For prisms or cylinders the 
areas of the upper, lower and middle planes are equal. 
The prismoidal formula when applied to the frustum of 
a pyramid saves the labor of extracting the square root, 
as required under the old rule. 



Mensuration for Sheet Metal Workers. 27 

As an example let us apply the above formula in ob- 
taining the solidity of the cone shown in Fig. 36. Follow- 
ing the above rule, we have as the area of the lower base 
314.16 and the area of the upper base o. Four times the 
area of the middle section, which is 10 inches in diameter, 
is 10 X 10 X 0.7854 = 78.54 X 4 = 314-16. The sum is 
314.16 + 314.16 = 628.32. One-sixth the perpendicular 
bight is 48 = 8; 8 X 628.32 = 5026.56 cubic inches. 

In Figs. 38 and 39 is shown the method of computing 
the volumes of the frustums of a square and hexagonal 
pyramid. Fig. 38 shows an inverted hopper, such as is 
usually made of heavy galvanized iron. Whether the 
shape is regular or irregular, the same method is used in 
determining the capacity. The top opening is 28 x 28 
inches, the bottom 10 x 10 inches, and the sides, a section 
taken midway between the top and bottom, would equal 

■ — ^ or 19 ; then 19 x 19 inches would be the size for 

the middle section, abed. The area of the upper plane 

equals 28 x 28 inches, or 784 square inches ; of the lower 

plane 10 x 10 inches, or 100 square inches, and of the 

middle plane 19 x 19 inches, or 361 square inches. Four 

times the middle plane is 1444. Then, following the rule, 

we have upper plane, 784, -{- lower plane, 100, -f four 

times middle plane, 1444, = 2328. This is multiplied by 

39 
one sixth the vertical bight, 39. Then 2328 X-^= I5'i32 

cubic inches. This divided by 231, the number of cubic 
inches in a gallon, gives 65 >4 gallons and i>^ cubic inches 
over. 

The same rule is applied to Figs. 39 to 42, inclusive. 
Fig. 39 shows a hopper hexagonal in shape, each side of 



28 Mensuration for Sheet Metal Workers. 

the lower base of which is 20 inches, the upper base 10 
inches, 

and a section taken midway between the two — 

or 15 inches, as a b. To obtain the area of these planes 
multiply the square of one of the sides of the regular poly- 



1,1,064,022.2208 ou. in. 




4,006 gallons 
and 5 gills 



Fig. 40. — Contents of Frustum of Cone. 

gon by the multiplier giyen in the following table for the 
proper polygon : 

Name. Sides to polygon. Multiplier. 

Triangle 3 0.433 

Square 4 1.000 

I'entagon 5 1.720 

Hexagon 6 2..598 

Heptagon 7 3.634 

Octagon 8 4.828 

Nonagon 9 6.182 

Decagon 10 7.694 

As the shape in question is a hexagon, we find the 

multiplier in the table to be 2.598. Then, following the 

above rule, the area of the lower plane is equal to 20 X 20 

==400; 400X2.598^1039.2. The area of the upper 

plane, 10 X 10 X 2.598 = 259.8, and four times the area 

of the middle plane, 15 X 15 X 2.598 X 4 = 2338.2. 

24 
Theii 1039.2 + 259.8 -f 2338.2 = 3637.2. 3637.2 X ^ = 

14,548.8 cubic inches contents. This divided by 231 = 63 
gallons, or 2 barrels, less 4.2 cubic inches. 



Mensuration for Sheet Metal Workers. 



29 



Fig. 40 shows the frustum of a cone. A tank of this 
form is usually made from ^^^-i^ch metal, riveted and re- 
inforced with angles and tees. The top diameter is 6 feet, 



bottom 10 feet and the middle diameter- 



10 



, or 8 feet. 



The area of the top section is 6 X 6 X 0.7854 = 28.2744 
square feet ; the bottom area, 10 X 10 X 0.7854 = 78.54 
square feet ; the middle area, 8 X 8 X 0.7854 = 50.2656. 
Then 28.2744 + 78.54 + (50.2656 X 4) = 307-8768 X 



12 



v-= 615.7536 cubic feet capacity. Multiply this amount 
by 1728, the number of cubic inches in a foot, and divide 




^ Contents 

5640 cu. In. 



Fig. 41. — Contents of Prismoid. 



k-14^ 



Contents 
0,76S cu. In. 



29 gallons 
.-* \ — ^X +69 ou. In 




Fig. 42. — Contents of Wedge. 



by 231, the number of cubic inches in a gallon, and we 
obtain the capacity in gallons. Thus, 615.7536 X 1728 = 
1,064,022.2208 cubic inches, which divided by 231 = 4606 
gallons and 5 gills. 

Fig. 41 shows an odd shaped vessel, whose parallel 
bases are right angle triangles. The bottom base is 20 x 
32 and contains 320 square inches ; the top 8 x 20, and 
contains 80 square inches. The size of the middle section 
is obtained by adding 20 and 32 and dividing by 2, which 
gives the length b c, or 26 inches. In similar manner 

--^ == 14, length of a b. Then the area oi a b c is. 



30 Mensuration for Sheet Metal Workers. 

^^ ^^ = t82. Then 80 + ( 182 X 4) + 320 = 1 128. 

30 
ii28X-^ = 5640 cubic inches. Divide this amount by 

231 and we get 24 gallons and 96 cubic inches over. 

In Fig. 42 is given another example in obtaining the 

capacity of odd shape solids, this example being in the 

form of a wedge. The rectangular base is 12 x 40 inches 

and contains in arear 480 square inches. The top runs to 

an apex 14 inches wide, whose area is o. A section taken 

midway between the top and bottom equals 6 X 27 inches 

and has an area of 162 square inches. Then o -|- (162 X 

06 
4) + 480 =1128. 1 1 28 X -^= 6768 cubic inches, 

6 

which, divided by 231=29 gallons, i quart and 11.25 

cubic inches over. 



Practical Examples for the Shop. 

The section of an 8 x 32 inch rectangular pipe is rep- 
resented in Fig. 43 of the diagrams. If a transition is 
made to a perfectly square pipe, what must each side 
measure? In solving this problem proceed as follows: 
Extract the square root of the area of the rectangular 
pipe, 256 square inches, which is 16 inches, the size of 
the square pipe. 

Supposing this 8 x 32 inch pipe was to form a transi- 
tion to another rectangular pipe, the width of which was 
12 inches, what must be its length to have the same area? 
Simply divide 256 by 12, and the quotient will be 21 1-3, 
making the size of the pipe 12 x 21 1-3 inches. 

If this 8 X 32 inch pipe were to form a transition to 
an oblong pipe with semicircular ends, 8 inches in diam- 



Mensuration for Sheet Metal Workers. 



31 



eter, as shown at A, what must the length of the distance 
be, shown from a to h? As two semicircles make a full 
circle, then deduct the area of the 8-inch circle from 256, 
and divide the remainder by eight, as follows : Area of 8- 
inch circle = 50.26; 256 — 50.26 = 205.74; 205.74-^ 8 = 
25.72, or 25% inches scant, the length from a to b. 

In Fig. 44 is shown a fitting in hot air piping known 
as a boot. With it a 14-inch round horizontal pipe and 
a vertical rectangular pipe whose width is 7 inches are 



y' 41!" — 



AREA 

256 
SQ. IN. 



\ 16 X 16 SQUARE 

V =12"x 21j' RECTANGULAR 

' s'x+lT ROUND ENDS, SEE A 



T 




Fig. 43. — Pipes Equal to Rectangular Pipe. 



Pig. 44.— Boot. 



joined. What must the length of this rectangular pipe be 
so that it will have the same area as the 14-inch round? 
The area of a 14-inch round pipe is 153.93, or approxi- 
mately 154 square inches. Divide 154 by 7, and the quo- 
tient will be 22 inches, the desired length. 

Suppose the size of the rectangular pipe is given, say 

7 X 22, and it is desired to know what size round pipe will 
have similar area? Divide the area of the given pipe, or 
154, by 0.7854, and extract the square root of the product. 
Thus 154 -^ 0.7854 =: 196. Vi96rri4. Then 14 inches 
is the diameter of the round pipe. 

In Fig. 45 is shown a chimney cap, which measures 

8 X 6^ inches at the bottom. The cap is to form a transi- 
tion from square to round. What must be the diameter 
at the top so that the area will be similar to the base? 



32 Mcnsnraiion for Sited Metal Workers. 

The same rule could be used as given in connection with 
Fig. 44, but by using a table of areas and circumferences 
of circles, much labor in computing can be saved. 

These tables are found in some text books, in engin- 
eers' pocket books and in the " Tinsmith's Helper and Pat- 
tern Book," published by the David Williams Company. 

In the following examples the use of these tables will 
be explained when applied to sheet metal work. Refer- 
ring to Fig. 46, the base measures 8 x 6^ inches and 
equals 50 square inches area. Instead of computing, to 



k- 8--J 



120.27 SQ. IN 





Fig. 45. — Chimney Top. Fig. 4G. — Boot. 

obtain the diameter of the top of the cap, simply refer to 
the table of areas and circumferences of circles, following 
the column under areas until 50.2655 is reached (the near- 
est to the required number 50), when we find it is the area 
of an 8-inch circle, which will be the top diameter of the 
cap. 

This method could have been used in Fig. 44, in which 
the area of the rectangular opening is 154 square inches. 
Follow the area column in the table until 153.9380 is 
reached, which suggests a 14-inch circle, as shown. 

In Fig. 46 is shown another form of boot. In this 
case the inlet A is 12^ inches in diameter and equals 



Mensuration for Sheet Metal Workers. 



33 



120.27 square inches. If the outlet B is to be oblong in 
shape with semicircular ends 6 inches in diameter, what 
must the length be of the straight side a h? Find the 
area of a 6-inch circle, which is 28.27 square inches. De- 
duct this from 120.27, which leaves 92. Divide this 
amount by six, the given width of the outlet, and the quo- 
tient will be 15 2-T) inches, the desired length of a b. Both 
sections then equal 120.27 square inches. 

In Figs. 47 and 48 is shown how to compute the 
amount that a duct must be increased in size in propor- 
tion to the number and size of inlets connected. Fig. 47 




AREA 
48 SQ. IN. 

i' I DEEP 

Fig. 47. — ripe with Branches. 



AREA 
50.27 SQ. IN. 



Fig. 48.— Pipe with Branches. 



shows a rectangular duct, whose depth is 6 inches 
throughout. Ducts of this kind are usually employed in 
ventilating work, and are increased in size every time a 
register is connected. A shows the first inlet, 6 x 15 
inches, having 90 square inches area. The second inlet, 
B, is 6 X 8 inches, and has 48 square inches area. The 
combined area of A and B is 138 square inches. Now 
what size must the duct be beyond B so as to have the 
area of the two inlets A and B? Divide 138 by 6 and 
the cjuotient will be 23, the width of the duct. This same 
rule is followed no matter how many inlets are taken up. 
In Fig. 48 is shown the method employed when the 
pipe is round. The first inlet is 14 inches and the second 
8. They contain, respectively. 153.93 and 50.27 square 
inches area. The combined area of the two inlets is 204.20 



34 



Mensuration for Sheet Metal Workers. 



square inches. The circle that has this area is found to 
be 16% inches in diameter. This is then the diameter of 
the large pipe. 

A triangular ventilating pipe is shown in Fig. 49. The 
size of the pipe is 16 x 32, containing one-half of the prod- 
uct of its dimensions, or 256 square inches area. What 
must the size of the various pipes be if a transition is 
desired to square, to rectangular 8 inches wide, and 
round ? To obtain the size of the square pipe, simply ex- 



-16- — J 




OF EQUAL AREA 
16"x 16" SQ. 
8"X 32" RECTANGLE 
ISnr ROUND 



Fig. 49. — Triangular Pipe. 




10 ROUND 

8. 862* SQUARE 

e'x 13.09° RECT. 



* I 

Fig. 50. — Elliptical Pipe. 



tract the square root of 256. Then 16 x 16 is the size of 
the square pipe. The length of the section of the rectang- 
ular pipe, whose width is desired to be 8 inches, is ob- 
tained by dividing 256 by 8. The quotient will be 32, or 
the required length. For the size of the round pipe, which 
should have an area equal to the triangular, follow the 
column of areas in the table of circle areas until 256.2398 
is reached, which will be found to equal the area of a cir- 
cle 18 1-16 inches in diameter and is the desired round 
pipe. 

In Fig. 50 is shown the section of an elliptical pipe 
measuring 8 x I2>^ inches. The area of this ellipse is 12.5 
X 8 X 0.7854 = 78.54 square inches. Suppose a transi- 
tion is to be made to round, square or rectangular pipe, 6 



Mensuration for Sheet Metal Workers. 



35 



inches wide, whose areas must be similar to the ellipse, 
what must their sizes be? Following the column of areas 
in the table, we find the area 78.54 is for a lo-inch round 
pipe. For the size of the side of a square pipe we have 
V78.54 = 8.862 -)- inches, or about 8% inches. Thus the 
square pipe would be 8% x 8% inches. For the length 
of the rectangular pipe, whose given width is 6 inches, 
divide 78.54 b}^ 6, and the quotient will be 13.09. Then 
6 X 13.09 inches will be the size of the rectangular pipe. 
In putting up ventilating, blower and blast pipes, it is 




AREA 194 SQ. IN. 

Fig. 51. — Two-Pronged Fork. 



A-B-C 

EACH 23r DIAM. 
COMBINED AREA 
1273.6 




1*. — 4or — >j 

AREA 1272.4 I 

Fig. 52. — Three-Pronged Fork. 



often the case that a number of branches are connected 
to one main, and the main pipe must have the combined 
area of the branches. An example of this is shown in 
Fig. 51, where two round branches are connected to a 
rectangular main pipe, whose width must be 121^ inches. 
What must the length of the section of this main pipe 
be to have a combined area of the two branches? The 
area of the lO-inch pipe is 78.54 square inches. The 
area of the i2i/^-inch pipe is 115.46 square inches. The 
sum of these two areas is 194 square inches ; 194 ^ 
12.125, the width of the main pipe, = 16. Therefore the 
size of the main pipe is I2>^ x 16 Fig. 52 shows three 



36 



Mensuration for SJicct Metal JVorkers. 



branches of round pipe, connecting in fork shape to a 
round main. Each of the branches, A, B and C, is 23^4 
inches diameter, the combined area of which equals 1273.6 
square inches. What must the size of the main be? Fol- 
lowing the table of areas, we find the nearest number to 
be 1272.4, which is the area of a pipe 4034 inches diame- 
ter. By using the tables of areas and circumference much 
time and labor are saved in computing. If these tables 






Fig. .53.— A Square Tank. 






Fig. 54.— A Sphere. Fig. .5.5.— An Oil Tank 



are not available, the calculations must be made by meth- 
ods described in the earlier installments of this series. 

Ascertaining Sizes of Articles. 

What follows is devoted to explaining how to obtain 
the unknown size of an article when the bight and capacity 
are given, or z'ice versa. The .first problem, shown in Fig. 
53, represents a water tank. Assume that a customer has 
ordered an 8-gallon tank, whose base is to measure 11 x 14 
inches. How high must it be to have the desired capacity ? 
The rule to follow in any square or rectangular tank is : 
Reduce the gallons to cubic inches; divide this amount by 
the area of the base, the quotient being the desired bight. 
As there are 231 cubic inches in the United States gallon, 



Mensuration for Sheet Metal Workers. 



37 



then in 8 gallons there will be 8 X 231 or 1848 cubic 
inches. The base is 11 x 14 and contains 154 square inch- 
es area. Then 1848-^-154=12 inches, the required 
hight of the tank. 

Suppose the hight and length of one of the sides and 
the capacity are given. What will the size of the remain- 
ing side be? Assuming the capacity to be 8 gallons, the 
hight 12 inches and the given side 11 inches, then divid- 
ing the number of cubic inches in 8 gallons by 12 and the 
quotient by 1 1 the result will be the required side. Thus, 




.:i_ 



Pig. 56. — Wash Boiler. Fig. 57. — Flaring Pail. 




Fig. 5S. — Flaring 
Measure. 



1848 -^ 12 = 154; 154 ^ II = 14, the required length of 
side. 

A copper ball Fig. 54, to be used as a float must con- 
tain 22,449 35-100 cubic inches. What must the diameter 
be to contain the above number of cubic inches ? The rule 
to follow is to divide the number of cubic inches by 
0.5236, and from the quotient thus obtained extract the 
cube root. As the sphere is to contain 22,449.35 cubic 
inches, then 22,449.35 ^ 0.5236 = 42,875. The next step 
is to extract the cube root of this quotient. 

When a number is multiplied by itself, as 5X5, the 
product, 25, is called the square of that number. When 



38 Mensuration for Sheet Metal Workers. 

a number is multiplied by itself twice, as 5 X 5 X 5, the 
product, 125, is called the cube of that number. There- 
fore the extraction of the cube root is nothing more than 
the finding of that number which, when multiplied by 
itself twice will result in the given number. To extract 
the cube root of 42,875, start at the decimal point and, 
counting to the left, separate the number into periods of 
three figures, as shown by 42'875. 

Cube 
Number, root. 
42'875.(35 
27 

Trial divisor. 

2700 15875 

42875 



Find the greatest number whose cube is contained in 
the first or left hand period, 42 ; 4 X 4 X 4 = 64, and is 
too great. Then take 3 ; 3 X 3 X 3 = 27. Therefore 3 is 
the first figure of the root. Subtracting 2y from 42 we 
obtain 15. Bring down the next period, 875, obtaining 
the first partial dividend, 15,875. Take three times the 
square of the root already found, which is 3- X 3 ^= 3 X 
3X3 =^2y. Annex two ciphers to it, and we have 2700 
for the trial divisor. Divide the trial divisor into 15,875, 
which suggests 5 as the second figure of the root. Prove 
this by multiplying 35 X 35 X 35, which equals 42,875, as 
shown, and leaves no remainder. Then 35 is the cube 
root of 42,875. Therefore the sphere in question must be 
35 inches in diameter. 

The trial divisor is sometimes contained in the partial 
dividend, a higher number than required. Whether it is 
too high or not can be ascertained by cubing the root 
found, and if its product is higher than the partial divi- 
dend a lower number must be taken, w'hose cube will be 
equal to or smaller than the partial dividend. If there 



Mensuration for Sheet Metal JJ'orkers. 39 

had been a remainder in the problem just shown, and it 
was desired to continue the root, periods of three ciphers 
each would have to be added to the whole number, 42,875, 
and continued as above described, so as to obtain the deci- 
mal part of the root. 

Tn Fig. 55 is shown an oil tank. Assume a tank whose 
diameter is 16 inches that must hold one barrel : Then 
what must be the hight of the tank ? Reduce the barrel to 
cubic inches, into which divide the area of the 16-inch 
circle. The quotient will be the required hight. One 
barrel equals 31^ gallons, or 7276.5 cubic inches. The 
area of a 16-inch circle is 201.062 square inches; 7276.5 -— 
201.062 =: 36.19 -[-. Therefore 36 1-5 inches is the de- 
sired hight. 

Suppose the tank is to hold the same quantity and the 
hight is to be 403/2 inches. What must the diameter be? 
In this case multiply the number of cubic inches by the 
hight and divide the quotient by 0.7854. From the quo- 
tient thus obtained extract the square root, which will be 
the desired diameter. 7276.5 cubic inches, the capacity of 
the tank, -f- 40.5 inches, the hight, = 179.6667; 179.6667 
jo.7854 = 228.758 -f ; V^8.758= 15.124, or 15^3 
inches, the desired diameter, as shown at B. 

In Fig. 56 is shown a lo-gallon wash boiler, whose 
base is 10 inches wide, with semicircular ends, the length 
of the straight part of the side from A to B being 954 
inches. What must be its hight? Reduce the capacity 
to cubic inches and divide it by the area of the bottom,. 
the quotient giving the required hight. Ten gallons equal 
2310 cubic inches. The area of the bottom equals the 
area of a lo-inch circle, which is 78.54, plus 92.5, the area 
of the rectangle 9^ X 10, = 171.04 square inches. 2310 
-r- 171.04 =. 13.5 inches, the hight of the boiler. 



40 Mensuration for Sheet Metal Workers. 

If for any reason the hight of the boiler and the diam- 
eter of the semicircular ends are given, and it is desired 
to know the width of the straight side A B, then find the 
capacity in cubic inches and divide by the hight ; from the 
quotient obtained subtract the area of the two semicir- 
cles and divide the remainder by the given width of the 
base, and the quotient will be the desired width. Thus: 
2310 ^ 13.5 = 171.11; 171-11 — 78-54 = 92.57; 92.57 
4- 10 = 9.26, the desired width of A B. 

In Figs. 57 and 58 are shown a flaring pail and a meas- 
sure whose capacities and top and bottom diameters are 
given, and it is required to find their hight. The rule ap- 
plicable to any form of flaring ware whose section is 
round, no matter what its capacity may be, is to find the 
number of cubic inches in the given capacity, which divide 
by the sum of the areas of the top and bottom diameters 
and four times the area of the middle section and multiply 
the quotient by six. As one quart contains 57.75 cubic 
inches, a lo-quart pail, shown in Fig. 57, will contain 
577.5 cubic inches. The area of the top diameter equals 
78.54, the bottom diameter 50.26, and the middle section, 
whose diameter is 9 inches, 63.61 square inches; 63.61 X 
4 = 254.44 square inches. Then 254.44 + 50.26 + 78.54 
= 383.24 ; 577.50 -:- 383.24 = 1.506 X 6 = 9.036 inches, 
the required hight. 

The measure shown in Fig. 58 is to hold one quart. 
Its top and bottom diameters are 2}^ and 4^4 inches, re- 
spectively. What must the hight be ? Following the same 
rule as above we have : Area of top equals 4.90. Area of 
bottom equals 14.18. The middle diameter equals 

£- 5 + 4.25 _ 2^2,7S' Its area equals 8.94. 8.94 X 4 = 
35.76. 



Mensuration for Sheet Metal Workers. 41 

Capacity of one quart equals 57.75 cubic inches ; di- 
vided by combined areas, or 54.84, leaves a quotient 1.053, 
which multiplied by 6 equals 6.318 inches, or hight. 

In Fig. 59 is shown the method of finding the hight 
in elliptical flaring ware when the top and bottom dimen- 
sions and capacity are given. The tub in this case is to 
hold 2i^ pints and 21^ cubic inches, the top dimensions 
to be II X 15)^ and the bottom 8 x I2>^ inches. What 
must the hight be? The rule to follow is the same as in 
the preceding problem. It should be remembered that 





Vig. 59. — Flaring Elliptical Tub. Fig. 60.— Flaring Pan. 

the area of an ellipse is found by multiplying the short 

and long diameters together and their product by 0.7854. 

Working this out, it will be found that 9 inches is the 

desired hight of the tub. 

Another problem where the same rule is employed is 

shown in Fig. 60, in which a drip pan to hold 29 quarts 

and 1314 cubic inches is illustrated. The top is to be 

16 X 22 and the bottom 12 x 18 inches. The middle sec- 

,• .• r , r „ 16 +12 18 4- 22 

tion'is found as follows: ! r=: 14- ^ =20 

2 ^2 

or 14 X 20 inches. 

Following the rule as before we have 6, the number 

of inches in the hight of the pan. 



42 Mensuration for Sheet Metal Workers. 

Short Rules in Computation. 

We now come to a point where a few short rules in 
computation may be of value to the sheet metal worker. 
In Fig. 6i let A B represent a wall with a rounded corner 
from a to b, on which a molding, gutter or cornice is to 
be placed, and it is desired to find the radius. To do this 
measure the distance from a to b, which is 5 feet. Bisect 
a b, obtaining d, and from d measure the distance, at 




Pig. <>1. — Finding Diametei' of Circle. 



right angles to a b, to the outside of the curve at e, which 

in this case measures i foot 3 inches. Divide the sum 

of the squares of one-half the chord and the rise by the 

rise, and one-half the cjuotient will be the desired radius. 

Thus, reducing to inches, we have, 

30-4- IS" Qoo-(-225 1 125 . . , 

-^^^ — ^—^ =^ ' ^= =375^ mches 

15X2 30 30 -^^^ 

or 3 feet i^ inches radius. 

In Fig. 62 is shov/n how the area of a given object 
can be obtained, even though it is so far away that meas- 
urements cannot be taken. This is obtained by propor- 



Mensuration for Sheet Metal Workers. 



43 



tion of triangles. Let us assume in this case that the 
spire shown at A is to be covered with metal or slate. If 
we obtain the contract, we will erect scaffolding to do the 
w^ork ; but it will not pay to erect scaffolding to take the 
measurements and measurements must be obtained to es- 
timate on the job. As we can get to the ridge of the 
roof, B C, the number of square feet in the four sides of 
the tower is obtained as follows : Measure a given dis- 




Fig. 62. — Finding Hight of Inaccessible Point. 

tance from the bottom of the roof of the spire, b — in 
this case 60 feet— to a. At a place a rod so high as to 
make the imaginary line b a horizontal. Take another 
rod and place it vertically at a distance of about 5 feet 
from a, as shown at C. Then, with the eye at a, cast the 
line of sight from a to the top of the spire, at d, and 
mark the second rod where the line of sight crosses it 
at C, and measure the distance from C on a c? to <? on 
a b, which in this case is 4 feet I'inch. Then, C <' is to 
d b 2^s c a \s io h a. Then the vertical hight from d to b 



44 



Mensuraiioii for SJicct Metal JVorkers. 



will be 49 feet. As the vertical liight is 49 feet the pitch 
d b will be a fraction of a foot more, but 49 will be safe 
enough for estimating purposes. 

In Fig. 63 is shown the method of finding the diam- 
eters of two circles whose areas are to be equal to a 
given area, or vice versa, and Fig. 64 shows various ex- 
amples solved b}- the steel square. Let A in Fig. 64 
represent two branches of 3-inch pipes, which are to be 





^-R 



< 5 6 7 8^9.1011 '213 141s I617iai9202 1 S2 S 1 

rM,l,l,l,M,l,MrffFm,l,i,l,lilil,l, 



7 ^ si 



Fig. 63. — Use of Steel Square. 



connected to a pipe whose area is equal to the areas of 
the two 3-inch pipes ; what must the diameter of the 
main pipe be? A? both the given pipes are 3 inches, then 
simply lay the 2-foot rule from 3 to 3 on the square, in 
Fig. 63, as shown by e f, which will measure 4^ inches, 
the diameter of the main pipe to have twice the area of 
one 3-inch pipe. 

Suppose the diameter of the main, or 4^-inch, pipe in 
Fig. 64 were given, and it was desired to take two 
branch pipes of equal diameter from it. Lay the rule 
of 434 inches over the angle of the square in Fig. 63 so 
that the distances from s to e and s to f are equal, then 

Lorc 



Mensuration for Sheet Metal IVorHers. 



45 



each side will measure 3 inches, as shown, the diameter 
of the branches shown in Fig. 64. 

In B are shown two branches, 7 and 12 inches, re- 
spectively; what must the diameter of the main be to 
equal their capacities? Set the rule on 7 and 12 in Fig. 
63, as shown, from a to b, which will measure 13^ inches, 
the diameter of the main pipe in B in Fig. 64. 

Again, suppose the main pipe of 13% inches diameter 
were given and a 7-inch pipe were already in place, what 
diameter must the other branch be so that the two 





Fig. 64. — Problems for Steel Square Solution. 

branches wih have the capacity of the main? Place the 
13^'8-inch mark of the rule on the 7-inch point, or a, and 
where the zero end intersects the lower edge of the square, 
which is at b, the 12-inch point on the square. This indi- 
cates that the diameter of the desired branch is 12 inches. 

Suppose in C in Fig. 64 we have a 16-inch pipe, to 
which a 4-inch branch was added. How much must the 
pipe increase in size? Place the rule on the square in 
Fig. 63 from 4 to 16, or from c to (/, and it will measure 
1 63/- inches, the size of the increased main in Fig. 64. 
This rule can also be used to advantage when square 
pipes are to be used. 

In Fig. 65 is shown the method of obtaining the 



^. V. 



46 



Mensuration for Sheet Metal Workers. 



length of radius by computation. This rule saves the 
time and labor of laying out the full size drawing of an 
article which has little flare and of obtaining the blank for 
a curved molding in cornice work, such as shown in the 
diagram A^ by a b. Let us suppose that a flaring collar is 
required, whose base, D C, is 40 inches in diameter, and 
top, A B, is 36 inches. How can the length of the radius 
be found ^ The rule to follow is to multiply the large 




^ ^i -s^^i'— 4/1 



/ 

/ 
1 
1 


/ 


\ 

9 eq.feei 




1 

e 




\ 


\ 


i! 


a 


ci I 




9 iq.teet 








«rea 


b 


.,^'' 





Fig. 65. — Obtaining Radius. 



—4-6-'!- J 

Fig. 66. — Rectangle to Square. 



diameter by the slant higlit and divide by the difference 
between the large and small diameters. The large diam- 
eter is 40 inches, the slant hight 8 inches, the small 
diameter 36 inches, and the difference between the large 

and the small is 4 inches. Then 8 X — = 80, or the 

length of the radius E C. 

The five following problems are devoted to obtaining 
sections of pipes of similar area to given sections by 
means of the compass and steel square. 



Mensuration for Sheet Metal Workers. 



47 



In Fig. 66 is shown how a rectangular section can be 
changed to a square section of eciual area. Let a b d 
represent a section 2 x 4.5 feet, which contains 9 square 
feet of area. Extend d a as d i, and with a as center 
and a ^ as radius, draw the quarter circle b c. Bisect 
c d zi e. With e as center draw the semicircle d c. 
JExtend b a until it intersects the circle at /; then, from f, 
complete the square / /; / a, which will be 3 x 3 feet and 
contain 9 square feet. 

Fig. 67 shows how a circle is changed to a square of 




— -5-5ir * 

Fig. 67. — Circle to Square. 




81 sq.in. 
area 



Fig. 08. — Three Squares. 



equal area. Divide the radius a b into four equal parts 
and place one of the parts from a to c and from d to e. 
Draw c e and complete the square c e f h, which w^ill 
measure, using the rule, 5 5-16 inches, the square of 
which will be its area and equal in area to the 6-inch cir- 
cle. 

Fig. 68 shows three separate square pipes, which are 
to be changed to one square pipe. One pipe is 2 x 2 
inches; the second, 6x6 inches, and the third, 9x9 
inches. Their total area is 121 square inches. In com- 
putation, we would simply extract the square root of 121 
and obtain the side. In this case, draw, in Fig. 69, the 
right angle a b c, making a b 2 inches and b c 6 inches. 



48 



Mensuration for Sheet Metal Workers. 



Draw a c, at right angles to which erect e d, g inches 
long. Draw d a and complete the square a. d c f, which 
will contain 121 square inches. 

In Fig. 70 is shown how to obtain a circle whose 
area is twice that of a given circle. Let A be the given 
circle ; draw the diameter a b, at right angles to which 
draw b e, equal to a b. Draw a c; bisect the same and ob- 



A(i 





Fig. 69.— Method of Fig. 68. Fig. 70.— Circle of Four Times the Area. 

tain d. With d as center and d a as radius draw the cir- 
cle B, which will contain twice the area of A. 

Finding Circumference of an Ellipse. 



The following communication, brought out by the 
publication of the articles in The Metal Worker, Plumber 
and Steam Fitter, refers to the rule for finding the cir- 
cumference of an ellipse : 

From Engineer, Neiv York. — In the article on " Men- 
suration for Sheet Metal Workers," the author gave a 
rule for finding the circumference of an ellipse that would 
only serve for the roughest kind of work. This rule — 
half the sum of the major and minor axes multiplied by 



Mensuration for Sheet Metal Workers. 49 

3.1416 — is fairly accurate when the elHpse approaches a 
circle — /. e., when the minor axis is nearly as long as 
the major axis. When the ellipse is very flat — /. e., when 
the minor axis is small, compared with the major axis, 
the formula is very inaccurate, and could not be suc- 
cessfully used in practice. This rule produces a result 
too small ; and for that reason should should not have 
been given for the use of the sheet metal worker, for 
it is a well-known principle in pattern cutting that the 
work, or the pattern, should not be made too small. If 
it cannot be made of the right size, it should be made too 
large, for it can always be cut down, but never can a 
small piece be cut larger. 

When D (the major axis) and d (the minor axis) 
are equal, the formula works out accurately ; when d is 
four-fifths of D, the error is — 0.42 per cent. ; when d is 
three-fifths of D, the error is — 1.76 per cent. ; when d is 
two-fifths of D, the error is — 4.60 per cent. ; when d is 
one-fifth of D, the error is — 10.45 P^i" cent. ; and when 
d equals o — /. e., when the ellipse has flattened to a 
straight line, the error is — 21.45 per cent. Below is 
given a table showing the workings of four formulas. 
In calculating the results shown, the major axis in each 
case has been assumed as 10, and the minor axis as 10, 
8, 6, 4, 2 and o. The first formula used, designated by 
A, is that given by the author of " Mensuration for Sheet 
Metal Workers," and which has just been referred to. 
It is as follows : 

Circumference = 3.1416 

After it, is shown the percentage of error at the sev- 
eral points. 

Formula B, the one next given, is : 



50 Mensuration for Slicct Metal JJ'orkcrs. 



Circumference = 3.1416 y ( ^ — ) 

This is sometimes simplified into the form : 
Circumference = 2.2215 ^ C-^" + ^' 
while this is not as simple to work out as A, from the 
column of errors that follows it one can see that it is 
much more accurate and that the error is in giving a 
somewhat too large result. 

Formula C, prepared by Trautwine for civil engineers, 
and which he claims is sufficiently accurate for ordinary 
principles, the error not exceeding more than i part in 
1000, when D does not exceed 5 d, is as follows : 

Circumference = '^.1416 i/ — ^^ — „ „ — 

^ ^ r 2 8.8 

If Z) exceeds five times d, then, instead of dividing 

(D — d)- by 8.8, divide it by the number in this table: 

D — Divide by D = Divide by 

6fZ 9. 20 d 9.8 

Id 9.2 • 2od 9.87 

8d 9.3 30 d 9.92 

9 d 9.35 40 d 9.98 

10 d 9.4 50 d 10.04 

12 d 9.5 60 d 10.10 

14 (Z 9.6 70 d 10.17 

16 d 9.68 80 d 10.2.3 

18 d 9.75 100 d 10.35 

Formula D is also quite accurate, and is somewhat 
neater to work than Formula C. It is as follows : 
Circumference =: 3. 1 416 d -\- 2(D — d) 

__ d(D — d) 

~ V (D + d) (D -\- 2d) 

In calculating the percentages of errors in Formulas 
A and C, it has been, assumed that Formula D is cor- 
rect, as it evidently is when d = D and when d ^= o. 



Mensuration for Sheet Metal Workers. 51 

The table comparing the formulas is as follows : 

^ = 10 Error. Error 

in o'i^-.o Percent. B. Percent. G. D 

10 31.42 31.42 31.42 3142 

8 28.27 - 0.42 28.45 + 0.21 28.37 28;39 

6 25.13 _ 1.76 25.91 + 1.29 25.57 25.58 

4 21.99 - 4.60 23.93 + 3.82 23.06 23 05 

2 18.85 -10.45 22.66 + 7.65 2102 -n'os 

15.71 -21.45 22.22 +11.10 19.96 2o'.00 

Sheet metal workers who have occasion to determine the 
circumference of the ellipse might do well to make note 
of these rules, because they will find it much quicker 
and cheaper to take a little more time in figuring out 
their work accurately, than to use a short formula with 
long error. 



JAM 28 1908 



